# Practice Exam I

Practice Exam I
Linear Algebra
September 21, 2004
1. Find a vector orthogonal to both 1,2,1 and  1,0,1 .
Let [ x, y , z ] be an arbitrary vector that is orthogonal to both of the given vectors.
So [ x, y, z ]  [1,2,1]  x  2 y  z  0 and [ x, y, z ]  [1,0,1]   x  z  0 . Now
solving the second equation we have that z = x and substituting this into the first
equation we have 2y = 0 or y = 0. Letting x = s where s is an arbitrary number
then z = s too. So [s, 0 ,s] is the general solution for a vector that is orthogonal to
the given two vectors and in particular [1,0,1] or [-1, 0 ,-1] are two of the infinite
vectors that would answer the problem.
2. Assume the 3x3 matrix B reduces to A by the following steps.
3  R3 3 R2
1  R2
2 4 R2
B R

 B1 R
 B2 R

 A
Give the elementary matrices whose product (in the proper order) with B gives A.
The elementary matrix that would work for the first row operation is
1 0 0
E1  0 1 0 . The elementary matrix that would work for the second row
0 3 1
1 0 0
operation is E 2  0 0 1 . The elementary matrix that would work for the last
0 1 0
1 0 0
row operation is E1  0 4 0 . So E3 E 2 E1 is the product of elementary
0 0 1
matrices whose product with B gives A.
3. Let u  [2,1,5] and v  [1,2,2]. Find the following.
(a) u
(b) u  v
(c) u  v
(d) 3u  2v
(e) The angle between u and v
(f) The distance between u and v
(a) u  (2) 2  1  5 2  30
(b) 6
2
(e)  68.58  or 1.20 radians
(f)
27
(d) [-8, 7, 11]
1 1
1 2
4. Find the inverse of the matrix 
0 1

0  1
0 0
0 0
1 1

1 0
 2 1
 1 1
Use algorithm we discussed in class to get the inverse to be 
 1 1

 2 2
0
0 
.
1

1  1
0
0
0
2 0 
0 3 1
1 2  1
5. Let A  1  3 , B  
, and C  
. Find the following if

 2 1  1
3 5  1


0 3 
possible, if not say why not.
(a) AB
(b) BA
0 6 2
(a)  6 0 4 


 6 3  3
(c) C  B
 3  6
(b) 

  3  6
(d) C  A
1 5 0 
(c) 

1 6  2
(d) can’t find this because the matrices C and A are of different sizes.
6. Solve the following system of equations using a matrix method.
3x  6 y  3z  0
y  2z  2
3x  7 y  6 z  2
3 6  3 0 


Set up the augmented matrix 0 1  2 2 and solve it to get the solution
3 7  6 2
x  [4,2,0] .
 1  2  1
7. Let A   2  4  2 . Give a basis for the null space, column space and row
 3 6
3 
space of the matrix A.
1  2  1
Row reduce the matrix down to 0 0
0  . The basis for the column space of

0 0
0 
A is {[1,2,3]} , the basis for the row space of A is {[1,-2,-1]}, and the basis for
the null space of A is {[2,1,0], [1,0,1]}.
8. For vectors u, v in  n prove that if u  v and u  v are perpendicular then
u  v .
Proof: if u  v and u  v are perpendicular then the dot product between them
must be 0. That is
(u  v)  (u  v)  u  (u  v)  v(u  v)  u  u  u  v  v  u  v  v  u  v  0  u  v .
The reason why we can cancel the middle dot products is because u  v  v  u .
 1  1
2 3
1
2
9. Let A 1  
and B 1  

. Find A and ( AB) .

1
2
1
1




 2  3
  1 4
and ( AB) 1  B 1 A 1  
A  2  A 1 A 1  

.
 3 5 
 0 1
10. Prove ( A  B)  C  A  ( B  C ) for all m x n matrices A, B, and C .
Proof: Since A, B, and C have the same size then (A+B)+C and A+(B+C) will
have the same size. We only need to prove that the (i,j)th entry of (A+B)+C is the
same as the (i,j)th entry of A+(B+C). So the (i,j)th entry of (A+B)+C is
(ai , j  bi , j )  ci , j  ai , j  (bi , j  ci , j ) (because of associativity property of real
numbers). The right side of this expression is the (i,j) th entry of A+(B+C) and
hence we are done.
11. Determine if {[1,3,2], [2,5,3], [4,0,1]} are linearly dependent or independent.
To determine if the vectors are linearly independent or dependent we must show
that r[1,3,2]  s[2,5,3]  t[4,0,1]  0 has a solution of [0,0,0] or a nonzero
solution. This is equivalent to solving the homogeneous system in matrix form
2 4
1
 3  5 0 . This is equivalent to showing the matrix row reduces to the


 2
3 1
identity or not. Thus if we row reduce the matrix, we will find out that it does row
reduced to the identity which says that the original vectors are independent.
2 3 1 
12. Given the matrix D  4  1 2 .


1 0 1
(a) Row reduce the matrix D to row echelon form.
(b) What is the rank of D?
(c) Solve Dx  0 .
(d) Determine the nullity of D.
(a)
(b)
(c)
(d)
1 1

1  4 2 
We get row echelon form is 0 1 0  .


0 0 1 


The rank of D is 3. That is rank(D)=3.
Using (a) we can solve the homogenous system. The solution is [ 0, 0, 0].
The nullity of D is the number of free variables in the solution in (c). Thus the
nullity(D) = 0.